我想要实现的是从 python 中的任何网站获取网站截图。
环境：Linux
【问题讨论】：
标签： python screenshot webpage backend
这是一个使用 webkit 的简单解决方案： http://webscraping.com/blog/Webpage-screenshots-with-webkit/
import sys
import time
from PyQt4.QtCore import *
from PyQt4.QtGui import *
from PyQt4.QtWebKit import *
class Screenshot(QWebView):
def __init__(self):
self.app = QApplication(sys.argv)
QWebView.__init__(self)
self._loaded = False
self.loadFinished.connect(self._loadFinished)
def capture(self, url, output_file):
self.load(QUrl(url))
self.wait_load()
# set to webpage size
frame = self.page().mainFrame()
self.page().setViewportSize(frame.contentsSize())
# render image
image = QImage(self.page().viewportSize(), QImage.Format_ARGB32)
painter = QPainter(image)
frame.render(painter)
painter.end()
print 'saving', output_file
image.save(output_file)
def wait_load(self, delay=0):
# process app events until page loaded
while not self._loaded:
self.app.processEvents()
time.sleep(delay)
self._loaded = False
def _loadFinished(self, result):
self._loaded = True
s = Screenshot()
s.capture('http://webscraping.com', 'website.png')
s.capture('http://webscraping.com/blog', 'blog.png')
【讨论】：
这是我从各种来源获取帮助的解决方案。它需要完整的网页屏幕截图并裁剪它（可选）并从裁剪的图像中生成缩略图。以下是要求：
要求：
npm -g install phantomjs
import os
from subprocess import Popen, PIPE
from selenium import webdriver
abspath = lambda *p: os.path.abspath(os.path.join(*p))
ROOT = abspath(os.path.dirname(__file__))
def execute_command(command):
result = Popen(command, shell=True, stdout=PIPE).stdout.read()
if len(result) > 0 and not result.isspace():
raise Exception(result)
def do_screen_capturing(url, screen_path, width, height):
print "Capturing screen.."
driver = webdriver.PhantomJS()
# it save service log file in same directory
# if you want to have log file stored else where
# initialize the webdriver.PhantomJS() as
# driver = webdriver.PhantomJS(service_log_path='/var/log/phantomjs/ghostdriver.log')
driver.set_script_timeout(30)
if width and height:
driver.set_window_size(width, height)
driver.get(url)
driver.save_screenshot(screen_path)
def do_crop(params):
print "Croping captured image.."
command = [
'convert',
params['screen_path'],
'-crop', '%sx%s+0+0' % (params['width'], params['height']),
params['crop_path']
]
execute_command(' '.join(command))
def do_thumbnail(params):
print "Generating thumbnail from croped captured image.."
command = [
'convert',
params['crop_path'],
'-filter', 'Lanczos',
'-thumbnail', '%sx%s' % (params['width'], params['height']),
params['thumbnail_path']
]
execute_command(' '.join(command))
def get_screen_shot(**kwargs):
url = kwargs['url']
width = int(kwargs.get('width', 1024)) # screen width to capture
height = int(kwargs.get('height', 768)) # screen height to capture
filename = kwargs.get('filename', 'screen.png') # file name e.g. screen.png
path = kwargs.get('path', ROOT) # directory path to store screen
crop = kwargs.get('crop', False) # crop the captured screen
crop_width = int(kwargs.get('crop_width', width)) # the width of crop screen
crop_height = int(kwargs.get('crop_height', height)) # the height of crop screen
crop_replace = kwargs.get('crop_replace', False) # does crop image replace original screen capture?
thumbnail = kwargs.get('thumbnail', False) # generate thumbnail from screen, requires crop=True
thumbnail_width = int(kwargs.get('thumbnail_width', width)) # the width of thumbnail
thumbnail_height = int(kwargs.get('thumbnail_height', height)) # the height of thumbnail
thumbnail_replace = kwargs.get('thumbnail_replace', False) # does thumbnail image replace crop image?
screen_path = abspath(path, filename)
crop_path = thumbnail_path = screen_path
if thumbnail and not crop:
raise Exception, 'Thumnail generation requires crop image, set crop=True'
do_screen_capturing(url, screen_path, width, height)
if crop:
if not crop_replace:
crop_path = abspath(path, 'crop_'+filename)
params = {
'width': crop_width, 'height': crop_height,
'crop_path': crop_path, 'screen_path': screen_path}
do_crop(params)
if thumbnail:
if not thumbnail_replace:
thumbnail_path = abspath(path, 'thumbnail_'+filename)
params = {
'width': thumbnail_width, 'height': thumbnail_height,
'thumbnail_path': thumbnail_path, 'crop_path': crop_path}
do_thumbnail(params)
return screen_path, crop_path, thumbnail_path
if __name__ == '__main__':
'''
Requirements:
Install NodeJS
Using Node's package manager install phantomjs: npm -g install phantomjs
install selenium (in your virtualenv, if you are using that)
install imageMagick
add phantomjs to system path (on windows)
'''
url = 'http://stackoverflow.com/questions/1197172/how-can-i-take-a-screenshot-image-of-a-website-using-python'
screen_path, crop_path, thumbnail_path = get_screen_shot(
url=url, filename='sof.png',
crop=True, crop_replace=False,
thumbnail=True, thumbnail_replace=False,
thumbnail_width=200, thumbnail_height=150,
)
这些是生成的图像：
【讨论】：
可以使用 Selenium
from selenium import webdriver
DRIVER = 'chromedriver'
driver = webdriver.Chrome(DRIVER)
driver.get('https://www.spotify.com')
screenshot = driver.save_screenshot('my_screenshot.png')
driver.quit()
https://sites.google.com/a/chromium.org/chromedriver/getting-started
【讨论】：
driver.execute_script("window.scrollTo(0, Y)") 滚动网页。其中“Y”是屏幕高度。您可以在循环中设置screenshot = driver.save_screenshot('my_screenshot.png') 和上面的代码，直到您的整个网页被覆盖。我对此不太确定，但这在逻辑上对我来说听起来不错。
driver.execute_script('document.body.style.zoom = "50%"')
在 Mac 上，有 webkit2png，在 Linux+KDE 上，你可以使用 khtml2png。我试过前者，效果很好，听说后者投入使用。
我最近遇到了QtWebKit，它声称是跨平台的（我猜是 Qt 将 WebKit 引入了他们的库）。但是我从来没有尝试过，所以我不能告诉你更多。
QtWebKit 链接显示了如何从 Python 访问。您至少应该能够使用 subprocess 对其他人做同样的事情。
【讨论】：
使用Rendertron 是一个选项。在底层，这是一个暴露以下端点的无头 Chrome：
/render/:url：访问这条路线，例如如果您对 DOM 感兴趣，请使用 requests.get。 /screenshot/:url：有兴趣截图的可以进入这条路线。您可以使用 npm 安装 rendertron，在一个终端中运行 rendertron，访问 http://localhost:3000/screenshot/:url 并保存文件，但在 render-tron.appspot.com 上提供了一个演示，可以在本地运行此 Python3 sn-p 而无需安装npm 包：
import requests
BASE = 'https://render-tron.appspot.com/screenshot/'
url = 'https://google.com'
path = 'target.jpg'
response = requests.get(BASE + url, stream=True)
# save file, see https://stackoverflow.com/a/13137873/7665691
if response.status_code == 200:
with open(path, 'wb') as file:
for chunk in response:
file.write(chunk)
【讨论】：
rendertron，然后它应该在端口 3000 上监听。然后，这个页面的截图应该可以在 @ 987654325@。您可以通过使用您最喜欢的浏览器浏览来检查，我的答案中的代码 sn-p 基本上只是将该图像存储到驱动器中。当然，您必须替换 BASE = 'http://localhost:3000/screenshot/' 和 url = 'https://stackoverflow.com/questions/1197172'。
11 年后...
使用Python3.6 和Google PageSpeedApi Insights v5 截取网站截图：
import base64
import requests
import traceback
import urllib.parse as ul
# It's possible to make requests without the api key, but the number of requests is very limited
url = "https://duckgo.com"
urle = ul.quote_plus(url)
image_path = "duckgo.jpg"
key = "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx"
strategy = "desktop" # "mobile"
u = f"https://www.googleapis.com/pagespeedonline/v5/runPagespeed?key={key}&strategy={strategy}&url={urle}"
try:
j = requests.get(u).json()
ss_encoded = j['lighthouseResult']['audits']['final-screenshot']['details']['data'].replace("data:image/jpeg;base64,", "")
ss_decoded = base64.b64decode(ss_encoded)
with open(image_path, 'wb+') as f:
f.write(ss_decoded)
except :
print(traceback.format_exc())
exit(1)
注意事项：
【讨论】：
我无法评论 ars 的答案，但实际上我使用 QtWebkit 运行了 Roland Tapken's code，它运行良好。
只是想确认 Roland 在他的博客上发布的内容在 Ubuntu 上运行良好。我们的生产版本最终没有使用他编写的任何内容，但我们使用 PyQt/QtWebKit 绑定取得了很大成功。
注意：以前的 URL 是：http://www.blogs.uni-osnabrueck.de/rotapken/2008/12/03/create-screenshots-of-a-web-page-using-python-and-qtwebkit/ 我已经用工作副本更新了它。
【讨论】：
这是一个老问题，大多数答案都有些过时了。 目前，我会做 2 件事中的 1 件事。
1.创建一个截取屏幕截图的程序
我会使用Pyppeteer 来截取网站的屏幕截图。这在 Puppeteer 包上运行。 Puppeteer 启动了一个无头 chrome 浏览器，因此屏幕截图看起来就像在普通浏览器中一样。
这取自 pyppeteer 文档：
import asyncio
from pyppeteer import launch
async def main():
browser = await launch()
page = await browser.newPage()
await page.goto('https://example.com')
await page.screenshot({'path': 'example.png'})
await browser.close()
asyncio.get_event_loop().run_until_complete(main())
2。使用截图 API
您还可以使用屏幕截图 API，例如 this one。 好处是您不必自己设置所有内容，只需调用 API 端点即可。
这取自屏幕截图 API 的文档：
import urllib.parse
import urllib.request
import ssl
ssl._create_default_https_context = ssl._create_unverified_context
# The parameters.
token = "YOUR_API_TOKEN"
url = urllib.parse.quote_plus("https://example.com")
width = 1920
height = 1080
output = "image"
# Create the query URL.
query = "https://screenshotapi.net/api/v1/screenshot"
query += "?token=%s&url=%s&width=%d&height=%d&output=%s" % (token, url, width, height, output)
# Call the API.
urllib.request.urlretrieve(query, "./example.png")
【讨论】：
您可以使用 Google Page Speed API 轻松完成您的任务。在我当前的项目中，我使用了用 Python 编写的 Google Page Speed API 查询来捕获提供的任何 Web URL 的屏幕截图并将其保存到某个位置。看看吧。
import urllib2
import json
import base64
import sys
import requests
import os
import errno
# The website's URL as an Input
site = sys.argv[1]
imagePath = sys.argv[2]
# The Google API. Remove "&strategy=mobile" for a desktop screenshot
api = "https://www.googleapis.com/pagespeedonline/v1/runPagespeed?screenshot=true&strategy=mobile&url=" + urllib2.quote(site)
# Get the results from Google
try:
site_data = json.load(urllib2.urlopen(api))
except urllib2.URLError:
print "Unable to retreive data"
sys.exit()
try:
screenshot_encoded = site_data['screenshot']['data']
except ValueError:
print "Invalid JSON encountered."
sys.exit()
# Google has a weird way of encoding the Base64 data
screenshot_encoded = screenshot_encoded.replace("_", "/")
screenshot_encoded = screenshot_encoded.replace("-", "+")
# Decode the Base64 data
screenshot_decoded = base64.b64decode(screenshot_encoded)
if not os.path.exists(os.path.dirname(impagepath)):
try:
os.makedirs(os.path.dirname(impagepath))
except OSError as exc:
if exc.errno != errno.EEXIST:
raise
# Save the file
with open(imagePath, 'w') as file_:
file_.write(screenshot_decoded)
不幸的是，以下是缺点。如果这些都不重要，您可以继续使用 Google Page Speed API。效果很好。
【讨论】：
使用网络服务 s-shot.ru（所以它不是那么快），但通过链接配置很容易设置需要的内容。 并且可以轻松截取整页截图
import requests
import urllib.parse
BASE = 'https://mini.s-shot.ru/1024x0/JPEG/1024/Z100/?' # you can modify size, format, zoom
url = 'https://stackoverflow.com/'#or whatever link you need
url = urllib.parse.quote_plus(url) #service needs link to be joined in encoded format
print(url)
path = 'target1.jpg'
response = requests.get(BASE + url, stream=True)
if response.status_code == 200:
with open(path, 'wb') as file:
for chunk in response:
file.write(chunk)
【讨论】：
你没有提到你在什么环境中运行，这会产生很大的不同，因为没有能够呈现 HTML 的纯 Python 网络浏览器。
但如果您使用的是 Mac，我使用 webkit2png 取得了巨大成功。如果没有，正如其他人指出的那样，有很多选择。
【讨论】：
我创建了一个名为 pywebcapture 的库，它封装了 selenium，它可以做到这一点：
pip install pywebcapture
使用 pip 安装后，您可以执行以下操作以轻松获得全尺寸屏幕截图：
# import modules
from pywebcapture import loader, driver
# load csv with urls
csv_file = loader.CSVLoader("csv_file_with_urls.csv", has_header_bool, url_column, optional_filename_column)
uri_dict = csv_file.get_uri_dict()
# create instance of the driver and run
d = driver.Driver("path/to/webdriver/", output_filepath, delay, uri_dict)
d.run()
享受吧！
【讨论】：
import subprocess
def screenshots(url, name):
subprocess.run('webkit2png -F -o {} {} -D ./screens'.format(name, url),
shell=True)
【讨论】：
webkit2png 的所有这些参数是什么？）以及文档链接。
webkit2png 默认不安装
试试这个..
#!/usr/bin/env python
import gtk.gdk
import time
import random
while 1 :
# generate a random time between 120 and 300 sec
random_time = random.randrange(120,300)
# wait between 120 and 300 seconds (or between 2 and 5 minutes)
print "Next picture in: %.2f minutes" % (float(random_time) / 60)
time.sleep(random_time)
w = gtk.gdk.get_default_root_window()
sz = w.get_size()
print "The size of the window is %d x %d" % sz
pb = gtk.gdk.Pixbuf(gtk.gdk.COLORSPACE_RGB,False,8,sz[0],sz[1])
pb = pb.get_from_drawable(w,w.get_colormap(),0,0,0,0,sz[0],sz[1])
ts = time.time()
filename = "screenshot"
filename += str(ts)
filename += ".png"
if (pb != None):
pb.save(filename,"png")
print "Screenshot saved to "+filename
else:
print "Unable to get the screenshot."
【讨论】：